Let $K/F$ be Galois with $G=Gal(K/F)$ and let $L$ be an intermediate field. Let $N\subseteq K$ be the normal closure of $L/F$. If $H=Gal(K/L)$ show that $Gal(K/N)=\cap_{\sigma\in G}\sigma H\sigma^{-1}$. (Exercise $8$, page $60$, Field and Galois Theory, Patrick Morandi.)

Help me a hint to prove it. Thanks a lot.

  • 2
    $\begingroup$ Show that $Aut_{\sigma(L)}(K) = \sigma Aut_L(K)\sigma^{-1}$ $\endgroup$ – Prahlad Vaidyanathan Nov 18 '13 at 9:30
  • $\begingroup$ Can you give me more details? $\endgroup$ – user109584 Nov 18 '13 at 9:34
  • $\begingroup$ Sorry I forgot to tag. $\endgroup$ – user109584 Nov 18 '13 at 9:48
  • $\begingroup$ Help me. please $\endgroup$ – user109584 Nov 19 '13 at 7:31
  • $\begingroup$ Have you been able to prove the hint I provided? $\endgroup$ – Prahlad Vaidyanathan Nov 19 '13 at 9:07

Here is my proof.

For convenience, let $M$ denote the group $\cap_{\sigma\in G} \sigma H\sigma^{-1}$. The notation $M(a)=a$ means that $a$ is fixed under the action of each element in $M$.

First we prove that $Gal(K/ N) \supseteq M$. It suffices to prove that $N$ is fixed by the group $M$. For any $a\in N$, since $N$ is the normal closure of $L/F$, there exists some $\sigma\in G$ such that $\sigma^{-1} (a) \in L$. Hence $H\sigma^{-1}(a)=\sigma^{-1}(a)$ because $L$ is fixed by $H$. Therefore, we see that $\sigma H\sigma^{-1}(a) = a,$ hence $M(a)=a$ because $M$ is contained in $\sigma H\sigma^{-1}$ by definition. Note that $a$ is arbitrary in $N$, hence $N$ is fixed by $M$.

Next we prove that $Gal(K/ N) \subseteq M$. Suppose $\tau\in Gal(K/N)$, for any $\sigma\in G$, $\tau\in\sigma H\sigma^{-1}\Leftrightarrow \sigma^{-1}\tau\sigma\in H\Leftrightarrow \sigma^{-1}\tau\sigma$ fixes $L$. So we only need to prove that $\sigma^{-1}\tau\sigma$ fixes $L$. $\forall a\in L$, we have $\sigma(a)\in N$ since $N$ is the normal closure of $L/F$. Because $\tau$ fixes $N$, we deduce that $\sigma^{-1}\tau\sigma(a) = \sigma^{-1}\sigma(a)=a$, which means that $\sigma^{-1}\tau\sigma$ fixes $L$, or $L$ is fixed by $\sigma^{-1}\tau\sigma$. We are done.


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